- #1

Jahnavi

- 848

- 102

## Homework Statement

## Homework Equations

Power of a bulb = I

^{2}R = V

^{2}/R

## The Attempt at a Solution

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From the ratings given on the bulbs using P=V

^{2}/R , we can conclude that resistances of the bulbs in decreasing order are R , P , Q i.e R has highest resistance and Q has lowest .

Now brightness is related to power dissipated .

Using P= I

^{2}R and the fact that both the current flowing in , as well as resistor of R is highest , so power dissipated in R should be highest .

Now potential difference across P and Q is same and resistance of P is more than that of Q , so using V

^{2}/R , power dissipated in P should be lower .

Hence least bright should be P . But R,P is not an option .

What is the mistake ?